Question

Let f be a function from $f:R\to R$ defined as $f\left( x \right)={{e}^{{{x}^{2}}}}+\cos x$ then f is
\[\begin{align}
  & \text{A}.\text{ One}-\text{one and onto} \\
 & \text{B}.\text{ One}-\text{one and into} \\
 & \text{C}.\text{ Many}-\text{one and onto} \\
 & \text{D}.\text{ Many}-\text{one and into} \\
\end{align}\]

Answer 1

Hint: First we will define one-one and onto function. If $g:X\to Y$ is a function then $g\left( {{x}_{1}} \right)=g\left( {{x}_{2}} \right)\Rightarrow {{x}_{1}}={{x}_{2}}$ then g is one-one otherwise many-one. So, we will assume a value $x,y\in R$ such that f (y) = f (x) then check whether x = y or not. Similarly, a function is onto if every element in the domain has a preimage, so, we will check if any value of k doesn't have pre image with respect to f in k.

Complete step by step answer:
Let us define a function $g:X\to Y$ such that \[g\left( x \right)=y\text{ }\forall \text{x}\in \text{X,y}\in \text{Y}\]
Then function g (x) is called one-one if for \[g\left( {{x}_{1}} \right)=g\left( {{x}_{2}} \right)\Rightarrow {{x}_{1}}={{x}_{2}}\] where ${{x}_{1}},{{x}_{2}}\in X$
If for \[g\left( {{x}_{1}} \right)=g\left( {{x}_{2}} \right)\Rightarrow {{x}_{1}}\ne {{x}_{2}}\] then g (x) is many-one function.
A function ‘g’ as defined above is called onto a function if \[\forall y\in Y, x\in X\] such that g (x) = y.
Basically every element $y\in Y$ has its pre image in $X\in Y$
If g (x) doesn't satisfy this condition then it is called into function.
We are given \[\begin{align}
  & f\left( x \right):R\to R \\
 & f\left( x \right)={{e}^{{{x}^{2}}}}+\cos x \\
\end{align}\]
Consider \[{{x}_{1}}=\pi \in R\text{ and }{{x}_{2}}=-\pi \in R\]
Then let us compute $f\left( \pi \right)\text{ and }f\left( -\pi \right)$
Value of \[f\left( \pi \right)={{e}^{{{\pi }^{2}}}}+\cos \pi \text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\] and value of $f\left( -\pi \right)$ is \[f\left( -\pi \right)={{e}^{{{\left( -\pi \right)}^{2}}}}+\cos \left( -\pi \right)\]
Now as \[\cos \left( -\theta \right)=\cos \theta \text{ and }{{\left( -\pi \right)}^{2}}=\pi \]
\[f\left( -\pi \right)={{e}^{{{\pi }^{2}}}}+\cos \pi \text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
From equation (i) and (ii) we see that
\[f\left( \pi \right)=f\left( -\pi \right)\text{ but }\pi \ne -\pi \]
For \[\pi \in R\text{ and }-\pi \in R\text{ and }f\left( \pi \right)=f\left( -\pi \right)\]
We did not get $\pi =-\pi $
Hence, f is not a one-one function.
Thus, f is a many-one function.
Finally we will check for onto.
f is defined from $R\to R$ which is real and has all positive and negative values. Now, consider the fact that ${{e}^{x}}\ge 0\text{ }\forall x\in 1R$
Graph of ${{e}^{x}}$ is as below:

So, \[f\left( x \right)={{e}^{{{x}^{2}}}}+\cos x\]
Let $x=\pi $ then as $\pi \in R$ so $f\left( \pi \right)$ is defined
$f\left( \pi \right)={{e}^{{{\pi }^{2}}}}+\cos \pi $
Now value of $\cos \pi =-1$
Now as \[{{e}^{{{\pi }^{2}}}}={{e}^{{{\left( 3.14 \right)}^{2}}}}\text{ and }{{\left( 3.14 \right)}^{2}}\text{ }>\text{ }1\]
By graph of ${{e}^{x}}$ we have \[{{e}^{{{\left( 3.14 \right)}^{2}}}}\text{ }>\text{ }1\]
As ${{e}^{0}}=1$

So as \[{{e}^{{{\pi }^{2}}}}\text{ }>\text{ }1\Rightarrow {{e}^{{{\pi }^{2}}}}-1\text{ }>\text{ 0}\]
Hence, $f\left( \pi \right)\text{ }>\text{ }0$ that is $f\left( \pi \right)$ is positive.
As a range of $\cos \theta =\left[ -1,1 \right]$ therefore, it can have no values less than -1 and greater than +1.
So, we have observed that the least value of $\cos x=-1$ at $x=\pi $ is giving $f\left( x \right)=f\left( \pi \right)$ as positive.
Therefore, as it is giving the value of f (x) at least the value of cos x.
i) For any values of cos x, f (x) is always positive. But f was defined from $R\to R$
ii) Negative reals are not in the image of f.
iii) There exist negative reals whose preimage is not present in f.
Therefore, f is not onto. f is into.

So, the correct answer is “Option D”.

Note: The possible confusion is the onto part. There is another way to see if the function f (x) is onto or not.
Observe that ${{e}^{x}}\text{ }>\text{ }0\text{ }\forall \text{x}\in \text{R}$ and $\cos \left( -\theta \right)=\cos \theta $ whose value lies between [-1, 1]. So, even if $\cos \theta $ has -1 as value then also the term ${{e}^{{{\pi }^{2}}}}-1$ is positive as ${{e}^{{{\pi }^{2}}}}={{e}^{{{\left( 3.14 \right)}^{2}}}}\text{ }>\text{ }1$